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2 years ago

Suppose a horse leans against a wall as in the figure below.

1 answer:

8 0

The force the horse and the rider exerts on the wall is equal to the weight combined acting in the opposite direction:

<h3>Force</h3>

Given Data

mass of horse and rider = 575kg

Force acting on wall = ??

When a body of mass rests on a surface, it exerts a force equal to the weight of the mass but opposite in direct on the mass/object

hence the force is computed as

Force = mass * acceleration

Force = 575 * 9.81

Force = 5640.75N

Learn more about force here:

brainly.com/question/12970081

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A 20 N unbalanced force causes an object to accelerate at 1.5 m/s2. What is the mass of the object?

Ket [755]

Answer:

Force = 20N

acceleration (a) = 1.5 m/s²

Mass of object (m) = ?

From Newtons II law

F = m. a N

m = F/a

m = 20/1.5

m = 13.34 Kg

Mass of an object is 13.34 Kg

8 0

3 years ago

Calculate the electric field strength at a point at which a test charge of 0.30 coulombs experiences a force of 5.0 newtons.

SVEN [57.7K]

The strength of electric field E is 17 N / C.

Explanation:

Electric field strength is defined as the force per unit charge acting at a point in the given field. The equation for the strength of the electric field is given by

E = F / q

where E represents the electric field strength,

F represents the force in newton,

q represents the charge in coulomb.

Given the charge q = 0.30 coulombs

force F = 5.0 N

Electric field strength E = force / charge

= 5.0 / 0.30

E = 16.66 = 17 N / C.

5 0

3 years ago

A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a

Burka [1]

Answer:

5 m/s, moving to the South.

Explanation:

Parameters given:

Mass of car, m = 1500 kg

Initial velocity of car, u = 15 m/s

Mass of truck, M = 4500 kg

Initial velocity of truck, v = 0 m/s (Truck is at rest)

Final velocity of car, U = 0 m/s (Car comes to a stop)

Final velocity of truck = V

Because the collision is elastic, we can apply the principle of conservation of momentum, we have that:

Total initial momentum = Total final momentum

m*u + M*U = m*v + M*V

(1500 * 15) + (4500 * 0) = (1500 * 0) + (4500 * V)

22500 + 0 = 0 + 00V

=> V = 22500/4500

V = 5 m/s

The velocity carries a positive sign, hence, it's moving in the same direction as the car was moving initially.

That is, it's moving to the South.

8 0

3 years ago

Read 2 more answers

Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge

Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃= 2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃